Question

p^2+q^2-2px-2qy+2xy by charpit's method find the complete integral this differential equation

Answer 1

1

Answer on Question #86005 – Math – Differential Equations

Question

$p^2 + q^2 - 2px - 2qy + 2xy$ by Charpit's method find the complete integral this differential equation.

Solution

Charpit's method is:

$$\frac {\mathrm {d} p}{\frac {\partial f}{\partial x} + p \frac {\partial f}{\partial z}} = \frac {d q}{\frac {\partial f}{\partial y} + q \frac {\partial f}{\partial z}} = \frac {d z}{- p \frac {\partial f}{\partial p} - q \frac {\partial f}{\partial q}} = \frac {d x}{- \frac {\partial f}{\partial p}} = \frac {d y}{- \frac {\partial f}{\partial q}}$$

and

$$d z = p d x + q d y,$$

where

$$f (x, y, z, p, q) = p ^ {2} + q ^ {2} - 2 p x - 2 q y + 2 x y,$$ $$\frac {\partial f}{\partial x} = - 2 p + 2 y, \frac {\partial f}{\partial y} = - 2 q + 2 x, \frac {\partial f}{\partial z} = 0, \frac {\partial f}{\partial p} = 2 p - 2 x, \frac {\partial f}{\partial q} = 2 q - 2 y.$$

Let's substitute (4) in (1):

$$\frac {\mathrm {d p}}{- 2 p + 2 y} = \frac {d q}{- 2 q + 2 x} = \frac {d z}{- p (2 p - 2 x) - q (2 q - 2 y)} = \frac {d x}{2 x - 2 p} = \frac {d y}{2 y - 2 q},$$

then take such fractions:

1) $\frac{\mathrm{dp}}{-2p + 2y} = \frac{dq}{-2q + 2x}$

$$\int \frac {\mathrm {d p}}{- 2 p + 2 y} = \int \frac {d q}{- 2 q + 2 x},$$ $$\ln | p - y | = \ln | q - x | + \ln c _ {1},$$ $$p - y = c _ {1} (q - x),$$ $$p = y + c _ {1} (x - q),$$ $$p = y + c _ {1} x - c _ {1} q.$$

2) $\frac{dx}{2x - 2p} = \frac{dy}{2y - 2q}$

$$\int \frac {d x}{2 x - 2 p} = \int \frac {d y}{2 y - 2 q},$$ $$\ln | x - p | = \ln | y - q | + \ln c _ {2},$$ $$x - p = c _ {2} (y - q),$$ $$p = x - c _ {2} y + c _ {2} q.$$

Let's substitute (5) in (6):

$$y + c _ {1} x - c _ {1} q = x - c _ {2} y + c _ {2} q,$$ $$q = \frac {(1 + c _ {2}) y + (c _ {1} - 1) x}{c _ {1} + c _ {2}}.$$

Let's substitute (7) in (5):

$$p = y + c _ {1} x - \frac {y + c _ {2} y + c _ {1} x - x}{c _ {1} + c _ {2}},$$ $$p = \frac {(c _ {1} + c _ {2}) y + c _ {1} (c _ {1} + c _ {2}) x - (1 + c _ {2}) y - (c _ {1} - 1) x}{c _ {1} + c _ {2}},$$ $$p = \frac {\left(c _ {1} - 1\right) y + \left(c _ {1} \left(c _ {1} + c _ {2}\right) - \left(c _ {1} - 1\right)\right) x}{c _ {1} + c _ {2}}.$$

Let's substitute (7), (8) in (2):

$$d z = \frac {(c _ {1} - 1) y + (c _ {1} (c _ {1} + c _ {2}) - (c _ {1} - 1)) x}{c _ {1} + c _ {2}} d x + \frac {(1 + c _ {2}) y + (c _ {1} - 1) x}{c _ {1} + c _ {2}} d y,$$ $$\int d z = \int \frac {(c _ {1} - 1) y + (c _ {1} (c _ {1} + c _ {2}) - (c _ {1} - 1)) x}{c _ {1} + c _ {2}} d x + \frac {(1 + c _ {2}) y + (c _ {1} - 1) x}{c _ {1} + c _ {2}} d y,$$ $$z = \frac {(c _ {1} - 1) y x + (c _ {1} (c _ {1} + c _ {2}) - (c _ {1} - 1)) x ^ {2}}{c _ {1} + c _ {2}} + \frac {(1 + c _ {2}) y ^ {2} + (c _ {1} - 1) x y}{c _ {1} + c _ {2}} + C,$$ $$z = \frac {(c _ {1} - 1) y x + (c _ {1} (c _ {1} + c _ {2}) - (c _ {1} - 1)) x ^ {2} + (1 + c _ {2}) y ^ {2} + (c _ {1} - 1) x y}{c _ {1} + c _ {2}} + C,$$ $$z = \frac {\left(c _ {1} (c _ {1} + c _ {2}) - (c _ {1} - 1)\right) x ^ {2} + (1 + c _ {2}) y ^ {2} + 2 (c _ {1} - 1) y x}{c _ {1} + c _ {2}} + C.$$

Answer: $z = \frac{(c_1(c_1 + c_2) - (c_1 - 1))x^2 + (1 + c_2)y^2 + 2(c_1 - 1)yx}{c_1 + c_2} + C$ is the complete integral of the differential equation $p^2 + q^2 - 2px - 2qy + 2xy$.

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